# Kinematic Equations: When & How to Use Each Formula (w/ Derivations)

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The kinematics equations describe the motion of an object undergoing constant acceleration. These equations relate the variables of time, position, velocity and acceleration of a moving object, allowing any of these variables to be solved for if the others are known.

Below is a depiction of an object undergoing constant acceleration motion in one dimension. The variable t is for time, position is x, velocity v and acceleration a. The subscripts i and f stand for "initial" and "final" respectively. It is assumed that t = 0 at xi and vi.

(Insert image 1)

## Kinematic Equations List

There are three primary kinematic equations listed below which apply when working in one dimension. These equations are:

\#\text{1: } v_f=v_i+at\\ \#\text{2: } x_f=x_i+v_i t+\frac 1 2 at^2\\ \#\text{3: }(v_f)^2 = (v_i)^2+2a(x_f - x_i)

## Notes About the Kinematic Equations

• These equations only work with a constant acceleration (which may be zero in the case of constant velocity).
• Depending upon which source you read, the final quantities may not have a subscript f, and/or might be represented in function notation as x(t) – read “x as a function of time” or “x at time t” – and v(t). Note that x(t) does NOT mean x multiplied by t!
• Sometimes the quantity xf - xi is written

Δx, meaning “the change in x,” or even simply as d, meaning displacement. All are equivalent. Position, velocity and acceleration are vector quantities, meaning they have direction associated with them. In one dimension, direction is typically indicated by signs – positive quantities are in the positive direction and negative quantities are in the negative direction.  Subscripts: "0" might be used for initial position and velocity instead of i. This "0" means "at t = 0," and x0 and v0 are typically pronounced "x-naught" and "v-naught." * Only one of the equations does not include time. When writing out givens and determining what equation to use, this is key!

## A Special Case: Free Fall

Free-fall motion is the motion of an object accelerating due to gravity alone in the absence of air resistance. The same kinematic equations apply; however, the acceleration value near the Earth’s surface is known. The magnitude of this acceleration is often represented by g, where g = 9.8 m/s2. The direction of this acceleration is downward, towards the Earth’s surface. (Note that some sources may approximate g as 10 m/s2, and others may use a value that is accurate to more than two decimal places.)

## Problem Solving Strategy for Kinematics Problems in One Dimension:

Sketch a diagram of the situation and choose an appropriate coordinate system. (Recall that x, v and a are all vector quantities, so by assigning a clear positive direction, it will be easier to keep track of signs.)

Write a list of known quantities. (Beware that sometimes the knowns aren’t obvious. Look for phrases like “starts from rest,” meaning that vi = 0, or “hits the ground,” meaning that xf = 0, and so on.)

Determine which quantity the question wants you to find. What is the unknown you will be solving for?

Choose the appropriate kinematic equation. This will be the equation that contains your unknown quantity together with known quantities.

Solve the equation for the unknown quantity, then plug in known values and compute the final answer. (Be careful about units! Sometimes you will need to convert units before computing.)

## One-Dimensional Kinematics Examples

Example 1: An advertisement claims that a sports car can go from 0 to 60 mph in 2.7 seconds. What is the acceleration of this car in m/s2? How far does it travel during these 2.7 seconds?

Solution:

(Insert Image 2)

Known and unknown quantities:

v_i=0\text{ mph}\\ v_f=60\text{ mph}\\ t=2.7\text{ s}\\ x_i=0\\ a=\text{?}\\ x_f=\text{?}

The first part of the question requires solving for the unknown acceleration. Here we can use equation #1:

v_f=v_i+at\implies a =\frac {(v_f-v_i)} t

Before we plug in numbers, however, we need to convert 60 mph to m/s:

60\cancel{\text{ mph}}\Bigg( \frac {0.477\text{ m/s}} {\cancel{\text{mph}}}\Bigg)=26.8\text{ m/s}

So the acceleration is then:

a=\frac {(26.8-0)} {2.7}=\underline{\bold{9.93}\text{ m/s}^2}

In order to find how far it goes in that time, we can use equation #2:

x_f=x_i+v_it+\frac 1 2 at^2=\frac 1 2 \times 9.93 \times 2.7^2=\underline{\bold{36.2}\text{ m}}

Example 2: A ball is thrown up at a speed of 15 m/s from a height of 1.5 m. How fast is it going when it hits the ground? How long does it take to hit the ground?

Solution:

(Insert image 3 )

Known and unknown quantities:

x_i=1.5\text{ m}\\x_f=0\text{ m}\\v_i=15\text{ m/s}\\a=-9.8\text{ m/s}^2\\v_f=?\\t=?

To solve the first part, we can use equation #3:

(v_f)^2=(v_i)^2+2a(x_f-x_i)\implies v_f=\pm \sqrt{(v_i)^2+2a(x_f-x_i)}

Everything is already in consistent units, so we can plug in values:

v_f=\pm \sqrt{15^2+2(-9.8)(0-1.5)}=\pm\sqrt{254.4}\approx\pm16\text{ m/s}

There are two solutions here. Which one is correct? From our diagram, we can see that the final velocity should be negative. So the answer is:

v_f=\underline{\bold{-16}\text{ m/s}}

To solve for time, we can use either equation #1 or equation #2. Since equation #1 is simpler to work with, we will use that one:

v_f=v_i+at\implies t=\frac {(v_f-v_i)} {a}=\frac {(-16-15)}{-9.8}\approx \underline{\bold{3.2}\text{ s}}

Note that the answer to the first part of the this question was not 0 m/s. While it is true that after the ball lands, it will have 0 velocity, this question wants to know how fast it is going in that split second before impact. Once the ball makes contact with the ground, our kinematic equations no longer apply because acceleration will not be constant.

## Kinematic Equations for Projectile Motion (Two Dimensions)

A projectile is an object moving in two dimensions under the influence of Earth's gravity. Its path is a parabola because the only acceleration is due to gravity. The kinematic equations for projectile motion take a slightly different form from the kinematic equations listed above. We use the fact that motion components that are perpendicular to each other – such as the horizontal x direction and the vertical y direction – are independent.

## Problem Solving Strategy for Projectile Motion Kinematics Problems:

Sketch a diagram of the situation. Just as with one-dimensional motion, it is helpful to sketch the scenario and indicate the coordinate system. Instead of using the labels x, v and a for position, velocity and acceleration, we need a way of labeling the motion in each dimension separately.

For the horizontal direction, it’s most common to use x for position and vx for the x-component of velocity (note that acceleration is 0 in this direction, so we don’t need a variable for it.) In the y direction, it is most common to use y for position and vy for the y-component of velocity. Acceleration can either be labeled ay or we can use the fact that we know the acceleration due to gravity is g in the negative y-direction, and just use that instead.

Write a list of known and unknown quantities by splitting the problem into two sections: vertical and horizontal motion. Use trigonometry to find the x- and y-components of any vector quantities that do not lie along an axis. It can be helpful to list this out in two columns:

(insert table 1)

Note: If velocity is given as a magnitude together with an angle, Ѳ, above the horizontal, then use vector decomposition, vx= vcos(Ѳ) and vy= vsin(Ѳ).

We can consider our three kinematic equations from before and adapt them to the x and y directions respectively.

X direction:

x_f=x_i+v_xt

Y direction:

v_{yf}=v_{yi}-gt\\ y_f=y_i+v_{yi} t-\frac 1 2 gt^2\\ (v_{yf})^2 = (v_{yi})^2-2g(y_f - y_i)

Note that the acceleration in the y direction is -g if we assume up is positive. A common misconception is that g = -9.8 m/s2, but this is incorrect; g itself is simply the magnitude of the acceleration: g = 9.8 m/s2, so we need to specify that the acceleration is negative.

Solve for one unknown in one of those dimensions, and then plug in what is common across both directions. While the motion in the two dimensions is independent, it happens on the same time scale, so the time variable is the same in both dimensions. (The time it takes the ball to undergo its vertical motion is the same as the amount of time it takes to undergo its horizontal motion.)

## Projectile Motion Kinematics Examples

Example 1: A projectile is launched horizontally from a cliff of height 20 m with an initial velocity of 50 m/s. How long does it take to hit the ground? How far from the base of the cliff does it land?

(insert image 4)

Known and unknown quantities:

(insert table 2)

We can find the time it takes to hit the ground by using the second vertical motion equation:

y_f=y_i+v_{yi} t-\frac 1 2 gt^2\implies t=\sqrt{\frac{(2\times 20)} g}=\underline{ \bold{2.02}\text{ s}}

Then to find where it lands, xf, we can use the horizontal motion equation:

x_f=x_i+v_xt=50\times2.02=\underline{\bold{101}\text{ s}}

Example 2: A ball is launched at 100 m/s from ground level at an angle of 30 degrees with the horizontal. Where does it land? When is its velocity the smallest? What is its location at this time?

(insert image 5)

Known and unknown quantities:

First we need to break the velocity vector into components:

v_x=v_i\cos(\theta)=100\cos(30)\approx 86.6 \text{ m/s}\\ v_{yi}=v_i\sin(\theta)=100\sin(30)=50 \text{ m/s}

Our table of quantities is then:

(insert table 3)

First we need to find the time the ball is in flight. We can do this with the second vertical equation_. Note that we use symmetry of the parabola to determine that the final _y velocity is the negative of the initial:

Then we determine how far it moves in the x direction in this time:

x_f=x_i+v_xt=86.6\times 10.2\approx\underline{\bold{883}\text m}

Using the symmetry of the parabolic path, we can determine that the velocity is smallest at 5.1 s, when the projectile is at the peak of its motion and the vertical component of velocity is 0. The x- and y-components of its motion at this time are:

x_f=x_i+v_xt=86.6\times 5.1\approx\underline{\bold{442}\text m}\\ y_f=y_i+v_{yi} t-\frac 1 2 gt^2=50\times5.1-\frac 1 2 9.8 \times 5.1^2\approx \underline{\bold{128}\text{ m}}

## Kinematic Equations Derivation

Equation #1: If the acceleration is constant, then:

a=\frac{(v_f-v_i)}{t}

Solving for the velocity, we have:

v_f=v_i+at

Equation #2: The average velocity can be written in two ways:

v_{avg}=\frac{(x_f-x_i)}{t}=\frac{(v_f+v_i)}{2}

If we replace _vf _with the expression from equation #1, we get:

\frac{(x_f-x_i)}{t}=\frac{((v_i+at)+v_i)}{2}

Solving for xf gives:

x_f=x_i+v_i t+\frac 1 2 at^2

Equation #3: Start by solving for t in equation #1

v_f=v_i+at \implies t=\frac{(v_f-v_i)}{a}

Plug this expression in for t in the average velocity relationship:

v_{avg}=\frac{(x_f-x_i)}{t}=\frac{(v_f+v_i)}{2}\implies \frac{(x_f-x_i)}{(\frac{(v_f-v_i)}{a})}=\frac{(v_f+v_i)}{2}

Rearranging this expression gives:

(v_f)^2 = (v_i)^2+2a(x_f - x_i)